The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500(mol-1)

  • A 54.0g
  • B 27.0g
  • C 13.5g
  • D 108.0g
jamb 2017chemistry

The correct answer is A. 54.0g

Recall that

mass deposited = \(\frac{MmIt}{96500n}\)

Mm =108, t = 4830s

I = 10A, n = 1

m = 108 × 10 × (\(\frac{4830}{96500}\)) × 1

m = 54.0g

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